# DC voltage follower

I'm working on a biomedical project, and need to transfer DC voltage from a source to a load. The problem is that the source has a high impedance relative to the load so I'm trying to have a DC voltage buffer (voltage follower) in between the source and the load. I came up with this circuit and seems that it works at least in simulation: https://www.circuitlab.com/circuit/xxkms467y7k8/dc-buffer/

As you can see in the figure, I don't want to use a separate VCC to power the V+ rail.

My questions are:

1- Would this circuit work in practice? I mean is it technically correct? Any suggestion to improve its performance? If not, what other options do I have to solve this problem? The solution in Passively Buffer Voltage didn't provide me with a low output impedance.

2- What are the best single supply op-amp options for this scenario? It has to handle a large range of v+ rail (from 1V to 15V). Also, the output impedance (RO) has to be low.

1- The load is the 100ohm resistor shown in the figure.

2- The output voltage must follow the input voltage (between 1 and 15V).

• I'm struggling to understand your thinking. You want to buffer the voltage so you can basically get more current, but you also want to source that current... from your high impedance source? That couldn't possibly work. Where is this extra power coming from? – BeB00 Mar 20 '18 at 21:33
• Be clear about output limits V and I (mA or A?) and flexible about input power and define load (MUST HAVE) if it is a bionic arm motor, that's a tall order – Sunnyskyguy EE75 Mar 20 '18 at 21:35
• If you had a perfect opamp which had 0ma of quiescent current, it seems like the output in your scenario should be 0.9V, so not sure whats going on with circuitlabs. I wouldnt rely on it for stuff more complicated than resistors and capacitors. – BeB00 Mar 20 '18 at 21:38
• Just to add, I tried it on partsim which uses pspice for simulations, and the simulation failed, so i guess it cant resolve the loop caused by the opamp output affecting its supply voltage. Regardless, this could never work. – BeB00 Mar 20 '18 at 21:48
• @user1512681 fundamentally no. You cannot create power (or rather, you cant create energy, but this is continuous so its the same thing). You need some other source of power. If you gave us more context around the device, there might be a different kind of solution, but I'm guessing probably not. – BeB00 Mar 20 '18 at 22:12

As pointed out in the comment - no, it will not work in practice. If 9V battery is going to be your source, then you need to use several in parallel.

• This answer could use an explanation. – Bort Mar 20 '18 at 22:00

Although Norms answer is correct, I feel like this needs more explanation.

First of all, what OP is clearly asking is:

Can I use my high impedance source to drive a low impedance output without losing voltage.

The answer is unequivocally NO. It doesnt matter what magic components you put in between, the problem is a basic issue of conservation of energy/power.

In op's example, he has a 9V source with 1k of series impedance. The maximum power that can possibly be extracted from this is if we match the source and load impedance together, by having a 1k load impedance. This results in:

P=IV=V^2/R=9*9/2k=40mW (then divide by 2 to get the power going into the load) = 20mW

This power output is a fundamental limitation, and whatever magic you do inbetween, with a 100 ohm load on the final output, the maximum voltage youll get is:

v=sqrt(P*R)=sqrt(0.04*100)=2V

This 2V is 7v short of what OP wants.

It is simply not possible without having another supply, and I'm not quite sure why other experienced posters are giving false hope.

edit: That last comment was in response to the other now deleted answers by other users.